poj 1742(好题，楼天城男人八题，混合背包)

Coins

Time Limit: 3000MS		Memory Limit: 30000K
Total Submissions: 33269		Accepted: 11295

Description

People in Silverland use coins.They have coins of value A1,A2,A3...An Silverland dollar.One day Tony opened his money-box and found there were some coins.He decided to buy a very nice watch in a nearby shop. He wanted to pay the exact price(without change) and he known the price would not more than m.But he didn't know the exact price of the watch.
You are to write a program which reads n,m,A1,A2,A3...An and C1,C2,C3...Cn corresponding to the number of Tony's coins of value A1,A2,A3...An then calculate how many prices(form 1 to m) Tony can pay use these coins.

Input

The input contains several test cases. The first line of each test case contains two integers n(1<=n<=100),m(m<=100000).The second line contains 2n integers, denoting A1,A2,A3...An,C1,C2,C3...Cn (1<=Ai<=100000,1<=Ci<=1000). The last test case is followed by two zeros.

Output

For each test case output the answer on a single line.

Sample Input

3 10
1 2 4 2 1 1
2 5
1 4 2 1
0 0

Sample Output

8
4

题意：题目给定 n种 硬币,第i种价值为 A[i],数量为C[i] ,问这些硬币能够组成1-m中哪些硬币。

分析：这题去年在hdu 2844上面过了，利用分组背包每次将每种物品分割，分割完一次马上进行就进行 01 背包，这样处理的速度就比所有的都分割完后再进行01背包要快许多。。但是没想到poj的数据如此变态。。然后看了网上的题解，恍然大悟。。原来当A[i]*C[i]>=m时，我们就可以将第i件物品看成无穷件了。。然后只要进行完全背包就OK。

hdu 2844 AC代码：

#include<stdio.h>
#include<algorithm>
#include<string.h>
#include<iostream>
using namespace std;

int  a[105],b[105],value[100000],f[100000];
int main()
{
    int n,m;
    while(scanf("%d%d",&n,&m)!=EOF&&n>0)
    {
        int wi=0;
        memset(f,0,sizeof(f));
        for(int i=0; i<n; i++)
        {
            scanf("%d",&a[i]);
        }
        for(int i=0; i<n; i++)
        {
            scanf("%d",&b[i]);
        }
        for(int i=0; i<n; i++)
        {
                int sum=1;
                int w=b[i],v=a[i];
                while(w>0)
                {
                    if(w>=sum)
                    {
                        value[wi]=sum*v;
                        w=w-sum;
                        sum=sum*2;
                    }
                    else
                    {
                        value[wi]=w*v;
                        w-=w;
                    }

                    for(int j=m; j>=value[wi]; j--)
                        f[j]=max(f[j],f[j-value[wi]]+value[wi]);
                        wi++;
                }



        }
        int ans=0;
        for(int i=1; i<=m; i++)
        {
           if(f[i]==i) ans++;
        }
        printf("%d
",ans);

    }
    return 0;
}

 poj 1742 AC代码:

#include<stdio.h>
#include<algorithm>
#include<string.h>
#include<iostream>
#define N 1005
#define M 100005
using namespace std;

int  A[105],C[105],W[N];
bool dp[M];
int main()
{
    int n,m;
    while(scanf("%d%d",&n,&m)!=EOF,n+m)
    {
        memset(dp,0,sizeof(dp));
        for(int i=1; i<=n; i++)
        {
            scanf("%d",&A[i]);
        }
        for(int i=1; i<=n; i++)
        {
            scanf("%d",&C[i]);
        }
        int k = 1;
        dp[0]=1;  ///初始化0元是存在的
        int ans = 0;
        for(int i=1; i<=n; i++)
        {
            if(C[i]*A[i]>=m) {  ///如果A[i]*C[i] >= m,我们可以将其当成完全背包问题处理，因为k*A[i]不能超过
                                ///m,超过了就可以看成"无限"了.
                for(int v = A[i];v<=m;v++){
                    if(!dp[v]&&dp[v-A[i]])  {
                        dp[v]=true;
                        ans++;
                    }
                }
            }else{
                int t = 1;
                while(C[i]>0){
                    if(C[i]>=t){
                        W[k]=A[i]*t;
                        C[i]-=t;
                        t = t<<1;
                    }else{W[k]=A[i]*C[i];C[i]=0;}
                    for(int v=m;v>=W[k];v--){
                        if(!dp[v]&&dp[v-W[k]])  {  ///当dp[v-W[k]]存在时才能够推导出dp[v]
                            dp[v]=true;
                            ans++;
                        }
                    }
                    k++;
                }
            }
        }
        printf("%d
",ans);

    }
    return 0;
}